# Potential in the schrödinger equation

by aaaa202
Tags: equation, potential, schrödinger
 P: 1,005 In clssical mechanics it doesn't really seem to matter which reference point you use for your potential. All that matters is the difference in potential between two points. Does the same hold true for the potential you plug into the Schrödinger equation? It doesn't seem so since plugging in V=-10J or V=0 should certainly provide different solutions. But the difference between the two potentials is merely just your choife of reference point, and that seems weird.
 Mentor P: 11,873 Have you solved the SE for the usual "particle in a box" example with V = 0 inside the box? If so, recall that the solution looks like $$\Psi_n(x,t) = \psi_n(x) e^{-iE_n t / \hbar}$$ Repeat the solution but set V = V0 inside the box. You should find that the solutions are $$\Psi_n(x,t) = \psi_n(x) e^{-i(E_n - V_0) t / \hbar}$$ with the same ##\psi_n(x)## as before. The solutions differ by an overall phase factor of ##e^{iV_0 t / \hbar}## which has no effect on probabilities and expectation values.
 P: 1,005 hmm okay makes sense. But as I recall the solution is something like psi(x)=Asin(kx). And k is related to the energy E-V0. So doesn't the energy determine k and thus does'nt the -V0 affect the sine term?
Mentor
P: 11,873
Potential in the schrödinger equation

 Quote by aaaa202 But as I recall the solution is something like psi(x)=Asin(kx). And k is related to the energy E-V0.
No matter what V0 is, you can still write ##\psi(x) = A \sin (n \pi x / L)##. All the energy levels get "bumped up" by V0, so the difference En - V0 remains the same regardless of V0.
 P: 1,005 okay so I believe you since the function of x is completely determined by the boundary conditions as you say. But is this true for all problems? I'm sorry, but I just don't see it that intuitively. Suppose you had a potential like the harmonic oscillator one: V = ½kx2 + V0. If you plug that into the Schrödinger equation, how can we then be sure that this additive constant doesn't alter the expectation values of different variables like x, p etc.
 Sci Advisor P: 5,464 You start with the time-dep. Schrödinger equation $$i\partial_t\,\psi_E(x,t) = [-\partial_x^2 + V(x)+V_0]\,\psi_E(x,t)$$ Now you make the usual ansatz $$\psi_E(x,t) = e^{-iEt} \,u_E(x)$$ You immediately get the time-indep. Schrödinger equation $$E\,\,u_E(x) = [-\partial_x^2 + V(x)+V_0]\,\,u_E(x)$$ And here you see that V0 can be absorbed into E $$E^\ast = E-V_0\;\to\;E^\ast\,\,u_{E^\ast}(x) = [-\partial_x^2 + V(x)]\,\,u_{E^\ast}(x)$$ The overall effect of varying V0 is - redefinition of the energy $E \to E^\ast$ - a global phase $\exp(-iV_0t)$ - the shape of the wave function $u_E(x)$ is not affected!! - therefore all physical quantities defined via time-independent operators $x, -i\partial_x, ...$ are not affected, either
 P: 65 Schrödinger equation is (excluding some constant that might be wrong...): $\frac{∂}{∂t}$=1/2h*(1/2m*p$^{2}$+qV) So, it tells you that the particle's phase in a point oscillates with a frequency proportional to its energy which is the sum of kinetic energy and potential. In stationary states (for example de 1-S state of the hydrogen atom), kinetic energy is different among different radius but the sum between this and the potential is the same and so all the points oscillate with the same phase and become an eigenstate. The phase to which the particle oscillates in the time is something you cannot measure and in fact you have a "gauge" in the base value of V. However, the way that changes V, ∇V, changes the relative phase of nearby point and that's equivalent to change the wave's momentum and therefore ∇V becomes a force. If you have a plane exp(jK·r) wave inside a region with a uniform difference of potential ∇V=dV•r, this wave will evolve its wave number K as long as the potential acts over it because the points where V is bigger will change its phase faster, changing the relative phase that was in the previous instant. I hope I could have been understood… Best regards, Sergio

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